Catch That Cow (经典广搜入门题)

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我是靠谱客的博主怕孤单航空，最近开发中收集的这篇文章主要介绍Catch That Cow (经典广搜入门题)，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Catch That Cow

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

题意简介：农夫追牛，牛不动，农夫有两种移动方式，第一种：前或后移动一步；第二种：农夫现在的位置X2；每次移动都消耗1分钟，求农夫追到牛的最短时间。

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#define MAXN 1000001
using namespace std;
int ret[MAXN],vis[MAXN];
queue<int> p;
int bfs(int n,int k)
{
int cur;
if(n==k) return 0;
p.push(n);
while(!p.empty()){
cur=p.front();
p.pop();
if(cur+1<MAXN && vis[cur+1]!=1){
ret[cur+1]=ret[cur]+1;
//步数加1
vis[cur+1]=1;
//标记是否已经读取过
p.push(cur+1);
//将这个符合条件的点放入队列
}
if(cur+1 == k) break;
//如果已经追上，则结束，并返回
if(cur-1>=0 && vis[cur-1]!=1){
//以下同上
ret[cur-1]=ret[cur]+1;
vis[cur-1]=1;
p.push(cur-1);
}
if(cur-1 == k) break;
if(cur*2<MAXN && vis[cur*2]!=1){
ret[cur*2]=ret[cur]+1;
vis[cur*2]=1;
p.push(cur*2);
}
if(cur*2 == k) break;
}
return ret[k];
}
int main()
{
int N,K;
cin>>N>>K;
printf("%dn",bfs(N,K));
return 0;
}